What is the solution for y^'-5y=(cos(x))*e^{5x} ?

The solution for y^'-5y=(cos(x))*e^{5x} is y=e^{5x}sin(x)+c_{1}e^{5x}

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y^'-5y=(cos(x))*e^{5x}

y^{^{'}} - 5 y = (cos (x)) \cdot e^{5 x}

y = e^{5 x} sin (x) + c_{1} e^{5 x}

Solution steps

y^{'} (x) - 5 y = (cos (x)) \cdot e^{5 x}

Solve linear ODE:

y = e^{5 x} sin (x) + c_{1} e^{5 x}

y = e^{5 x} sin (x) + c_{1} e^{5 x}

\frac{1}{2} y^{^{'}} + y = 0

\frac{d}{d x} (sin (x - 1))

\int cos (t) d t

\int (- 5 x^{6} - 3 x^{3} - 6 x + 5) d x

\int_{- 1}^{1} (e^{x}) - (x^{2} - 1) d x

What is the solution for y^'-5y=(cos(x))*e^{5x} ?
The solution for y^'-5y=(cos(x))*e^{5x} is y=e^{5x}sin(x)+c_{1}e^{5x}